$G$ acts properly discontinuous upon $\mathbb{R}$ so $G\cong\mathbb{Z}$

Question

This question was asked here: If the action of a group $G$ on $\mathbb{R}$ is properly discontinuous then G is isomorph to $\mathbb{Z}$? but I wish to get a different point of view for a solution. I hope it is not considered as a duplicate.

I took the next approach for answering this question, assuming that $G$ is not a trivial group ($|G|=1$).

I proved the next claim: If $P:X\rightarrow Y$ is a covering map and $X$ is a $n$ dimensional manifold, then $Y$ is also a $n$ dimensional manifold.

Using the theorem that in case $G$ acts PD upon $X$, $P:X\rightarrow X/G$ defined as $P(x)=O(x)$ is a covering space and $G\cong \Delta$ when $\Delta$ is the set of deck transformations, we get from the previous claim and the fact the $\mathbb{R}$ is 1 dimensional manifold that $X/G$ is a one dimensional manifold means $X/G$ homeomorphic to $[a,b]$ or $\mathbb{R}$ or $S^1$. I want to explain that is must be homeomorphic to $S^1$ and therefore $G$ must be homeomorphic to $\mathbb{Z}$ because $S^1 \cong \mathbb{R}/\mathbb{Z}$.

So for $[a,b]$, $P$ a covering space is a local homeomorphism, so every $x\in \mathbb{R}$ is an interior point $\Rightarrow$ $P(x)$ is an interior point, therefore $\forall x\in \mathbb{R}/G$ $x$ is an interior point $\Rightarrow$ $\mathbb{R}/G$ can't be homeomorphic to $[a,b]$.

For $\mathbb{R}$ in case $\mathbb{R}/G \cong \mathbb{R}$ we conclude that $\pi_1(\mathbb{R}/G)$ is trivial (cause $\pi_1(\mathbb{R})$ is trivial), an other theorem gives us that when $P:X\rightarrow Y$ is a cover projection and $X$ is simply connected ($\pi_1(X)$ is trivial) then $\pi_1(Y) \cong \Delta$ so under this case we get that $G\cong \Delta \cong \pi_1(\mathbb{R}/G)$ are all trivial groups, but $G$ is not trivial.

So all we are left with is $R/G \cong S^1$ as we wanted.

Answer 1

Note that if $G$ acts properly discontinuously on $\mathbb{R}$, then it acts properly and freely on it; hence you already know that $\mathbb{R}/G$ is a manifold without boundary. You also know that it's connected, hence it's $S^1$ or $\mathbb{R}$ (because it's $1$-dimensional)

You can then go through your proof and prove that $\mathbb{R}/G\simeq S^1$. And then you say "and because $\mathbb{R/Z}\simeq S^1$, we must have $G\simeq \mathbb{Z}$.

This is a bit shaky. Indeed it seems you conclude this without any work, and so it would be true for any example without any hypotheses on the action, but it's not; and you can figure out examples for yourself. Here's a good solution (hidden at first, if you want to keep looking for yourself)

Because $G$ acts properly discontinuously on $\mathbb{R}$, the projection $p$ is a covering map, hence we have a covering $\mathbb{R}\to S^1$. Now since $\mathbb{R}$ has trivial $\pi_1$, this is (by the classification theorem) isomorphic to the usual covering $e: t\mapsto e^{it}, \mathbb{R}\to S^1$, hence they have the same automorphism group, $\mathbb{Z}$. But the automorphism group of $\mathbb{R}\to \mathbb{R}/G$ is $G$, so $G\simeq\mathbb{Z}$